6.7 Probability and Sets

A Small Area With Reliable Traps

Probability and sets: Calculating probability (simple, compound, and conditional), and defining sample spaces and events using set notation.

At 1 to 3 of 20 questions this is one of the smaller QAS areas, but its error rate is high because the traps are conceptual rather than computational.

Sets and Sample Spaces

A set is a collection of distinct elements, written in braces: $A = {2, 4, 6, 8}$.

The sample space $S$ is the set of all possible outcomes. An event is a subset of $S$.

Rolling one die: $S = {1, 2, 3, 4, 5, 6}$, and "rolling even" is the event $E = {2, 4, 6}$.

Set Operations

With $A = {1,2,3,4}$ and $B = {3,4,5,6}$:

• $A \cup B = {1,2,3,4,5,6}$
• $A \cap B = {3,4}$

Note that union does not double-count 3 and 4. Sets contain distinct elements, and this non-duplication is exactly why the probability addition rule subtracts an overlap.

Simple Probability

$P(E) = \frac{\text{number of favorable outcomes}}{\text{total number of equally likely outcomes}}$

Every probability satisfies $0 \le P(E) \le 1$. An answer choice outside that interval can be eliminated on sight.

A bag holds 5 red, 3 blue, and 4 green marbles. P(blue)? Total $= 12$. $P(\text{blue}) = \dfrac{3}{12} = \dfrac{1}{4}$

The Complement Rule

$P(\text{not } E) = 1 - P(E)$

This is the single most useful shortcut in the area. Whenever a question says "at least one," compute the complement:

Flip a fair coin 3 times. P(at least one head)?

Directly, you would count 7 favorable outcomes out of 8. Via the complement:

$P(\text{no heads}) = P(TTT) = \left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$ $P(\text{at least one head}) = 1 - \tfrac{1}{8} = \tfrac{7}{8}$

Compound Probability

Independent Events — Multiply

Events are independent when the first outcome does not change the second's probability.

$P(A \text{ and } B) = P(A) \cdot P(B)$

Two fair dice. P(both show 4)? $\tfrac{1}{6} \cdot \tfrac{1}{6} = \tfrac{1}{36}$

Dependent Events — With and Without Replacement

This is where most errors occur.

A bag holds 5 red and 3 blue marbles. Two are drawn.

With replacement (first marble returned):

$P(\text{both red}) = \tfrac{5}{8} \cdot \tfrac{5}{8} = \tfrac{25}{64}$

Without replacement (first marble kept):

$P(\text{both red}) = \tfrac{5}{8} \cdot \tfrac{4}{7} = \tfrac{20}{56} = \tfrac{5}{14}$

On the second draw both numbers change: one red is gone (5 → 4) and one marble total is gone (8 → 7). The classic error updates only the numerator, giving $\tfrac{5}{8} \cdot \tfrac{4}{8}$.

Signal words: without replacement, does not replace, keeps, sets aside, eats. Also, any scenario drawing people from a group is inherently without replacement — the same person cannot be chosen twice.

The Addition Rule

$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$

The subtraction removes the double-counted overlap.

Draw one card from a standard 52-card deck. P(heart or face card)?

• $P(\text{heart}) = \tfrac{13}{52}$
• $P(\text{face card}) = \tfrac{12}{52}$ (J, Q, K in four suits)
• $P(\text{both}) = \tfrac{3}{52}$ (J, Q, K of hearts)

$P = \tfrac{13}{52} + \tfrac{12}{52} - \tfrac{3}{52} = \tfrac{22}{52} = \tfrac{11}{26}$

Skipping the subtraction gives $\tfrac{25}{52}$ — a distractor that will be offered.

When events are mutually exclusive (cannot both occur), the overlap is 0 and the rule reduces to simple addition.

Conditional Probability

$P(B \mid A)$ is the probability of B given that A has occurred. Conditioning restricts the sample space to the outcomes where A is true.

Of 40 students, 22 take Spanish, 18 take chemistry, and 10 take both. Given that a student takes Spanish, what is the probability they take chemistry?

Restrict to the 22 Spanish students. Of those, 10 also take chemistry:

$P(\text{chem} \mid \text{Spanish}) = \dfrac{10}{22} = \dfrac{5}{11}$

The denominator is 22, not 40. Using the full group is the defining error in conditional probability, and it is what the item is written to detect.

Two-Way Tables

Many items present a table. Read the denominator from the question's condition:

• $P(\text{passed})$ → denominator 70 → $\tfrac{55}{70} = \tfrac{11}{14}$
• $P(\text{passed} \mid \text{evening})$ → denominator 30 → $\tfrac{21}{30} = \tfrac{7}{10}$
• $P(\text{evening} \mid \text{passed})$ → denominator 55 → $\tfrac{21}{55}$

The last two use the same cell with different denominators. The word after "given" names the denominator.